# Integration by Partial Fractions – Integrals

If f(x) and g(x) are polynomial functions such function. that g(x) ≠ 0 then f(x)/g(x) is called **Rational Functions**. If degree f(x) < degree g(x) then f(x)/g(x) is called a **proper rational function**. If degree f(x) < degree g(x) then f(x)/g(x) is called an **improper rational function**. If f(x)/g(x) is an improper rational function then by dividing f(x) by g(x), we can express f(x)/g(x) as the sum of a polynomial and a proper rational function.

**Partial Fractions**

Any proper rational function p(x)/q(x) can be expressed as the sum of rational functions, each having the simplest factor q(x), each such fraction is known as a partial function and the process of obtaining them is called the decomposition or resolving of the given function into partial fractions.

### Integration** by Partial Fractions**

For example lets say we want to evaluate ∫[p(x)/q(x)] dx where p(x)/q(x) is in a proper rational fraction. In cases like these, we can write the integrand as in a form of the sum of simpler rational functions by using partial fraction decomposition after that integration can be carried out easily. Here The values of A, B, C, etc. can be obtained as per the question.

Factors in the denominator or denominators in rational functions | Corresponding Partial Fractions |
---|---|

(x – a) | A/(x – a) |

(x – b) | A/(x – b) + A/(x – b) |

(x – c) | A/(x – c) + B/(x – c) |

(ax | Ax + B/(ax |

### Examples

**Example 1: Evaluate ∫(x – 1)/(x + 1)(x – 2) dx?**

**Solution:**

Let (x – 1)/(x + 1)(x – 2) = A/(x + 1) + B/(x – 2)

Then, (x – 1) = A(x – 2) + B(x + 1) ………………(i)

Putting x = -1 in (i), we get A = 2/3

Putting x = 2 in (i), we get B = 1/3

Therefore,

(x – 1)/(x + 1)(x – 2) = 2/3(x + 1) + 1/3(x – 2)

=> ∫(x – 1)/(x + 1)(x – 2) = 2/3∫dx/(x + 1) + 1/3∫dx/(x – 2)

= 2/3log | x + 1 | + 1/3 log | x – 2 | + C

**Example 2: Evaluate ∫dx/x{6(log x) ^{2 }+ 7log x + 2}?**

**Solution: **

Putting log x = t and 1/x dx = dt, we get

I = ∫dx/x{6(log x)

^{2}+ 7log x + 2} = ∫dt/(6t^{2}+ 7t + 2) = ∫dt/(2t + 1)(3t + 2)Let 1/(2t + 1)(3t + 2) = A/(2t + 1) + B/(3t + 2)

Then, 1 ≡ A(3t + 2) + B(2t + 1) …………………….(i)

Putting t = -1/2 in (i), we get A = 2

Putting t = -2/3 in (i), we get B = -3

Therefore, 1/(2t + 1)(3t + 2) = 2/(2t + 1) + (-3)/(3t + 2)

=> I = ∫dt/(2t + 1)(3t + 2)

= ∫2dt/(2t + 1) – ∫3dt/(3t – 2)

= log | 2t + 1 | – log | 3t + 2 |

= log | 2t + 1 |/log | 3t + 2 | + C

= log | 2 log x + 1 | / log | 3 log x + 2 | + C

**Example 3: Evaluate ∫dx/(x ^{3 }+ x^{2 }+ x + 1)?**

**Solution:**

We have 1/(x

^{3 }+ x^{2 }+ x + 1) = 1/x^{2}(x + 1) + (x + 1) = 1/(x + 1)(x^{2 }+ 1)Let 1/(x + 1)(x

^{2 }+ 1) = A/(x + 1) + Bx + C/(x^{2 }+ 1) ……………………(i)=> 1 ≡ A(x

^{2 }+ 1) + (Bx + C) (x + 1)Putting x = -1 on both sides of (i), we get A = 1/2.

Comparing coefficients of x

^{2}on the both sides of (i), we getA + B = 0 => B = -A = -1/2

Comparing coefficients of x on the both sides of (i), we get

B + C = 0 => C = -B = 1/2

Therefore, 1/(x + 1) (x

^{2 }+ 1) = 1/2(x + 1) + (-1/2x + 1/2)/(x^{2 }+ 1)Therefore, ∫1/(x + 1) (x

^{2 }+ 1) = ∫dx/(x + 1) (x^{2 }+ 1)= 1/2∫dx/(x + 1) – 1/2∫x/(x

^{2 }+ 1)dx + 1/2∫dx/(x^{2 }+ 1)= 1/2∫dx/(x + 1) – 1/4∫2x/(x

^{2 }+ 1)dx + 1/2∫dx/(x^{2 }+ 1)= 1/2 log | x + 1 | – 1/4 log | x

^{2}+ 1 | + 1/2 tan^{-1}x + C

**Example 4: Evaluate ∫x ^{2}/(x^{2} + 2)(x^{2 }+ 3)dx?**

**Solution: **

Let x

^{2}/(x^{2}+ 2) (x^{2}+ 3) = y/(y + 2)(y + 3) where x^{2 }= y.Let y/(y + 2) (y + 3) = A/(y + 2) + B/(y + 3)

=> y ≡ A(y + 3) + B/(y + 2) ………………(i)

Putting y = -2 on the both sides of (i), we get A = -2.

Putting y = -3 on the both sides of (i), we get B = 3.

Therefore, y/(y + 2) (y + 3) = -2/(y + 2) + 3/(y + 3)

=> x

^{2}/(x^{2 }+ 2) (x^{2 }+ 3) = -2/(x^{2 }+ 2) + 3/(x^{2 }+ 3)=> ∫x

^{2}/(x^{2 }+ 2) (x^{2 }+ 3) = -2∫dx/(x^{2 }+ 2) + 3∫dx/(x^{2 }+ 3)= -2/√2tan

^{-1}(x/√2) + 3/√3tan^{-1}(x/√3) + C= -√2tan

^{-1}(x/√2) + √3tan^{-1}(x/√3) + C

**Example 5: Evaluate ∫dx/x(x ^{4 }+ 1)?**

**Solution:**

We have

I = ∫dx/x(x

^{4 }+ 1) = ∫x^{3}/x^{4 }(x^{4 }+ 1) dx [multiplying numerator and denominator by x^{3}].Putting x

^{4}= t and 4x^{3}dx = dt, we getI = 1/4∫dt/t(t + 1)

= 1/4∫{1/t – 1/(t + 1)}dt [by partial fraction]

= 1/4∫1/t dt – 1/4∫1/(t + 1)dt

= 1/4 log | t | – 1/4 log | t + 1 | + C

= 1/4 log | x

^{4}| – 1/4 log | x^{4}+ 1 | + C= (1/4 * 4) log | x | – 1/4 log | x

^{4}+ 1 | + C= log | x | – 1/4 log | x

^{4}+ 1 | + C

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